Question

Chickens have different comb morphologies which can be walnut, pea, single or rose. Suppose that a walnut combed hen is crossed with a single combed roster; the F1 progeny all have walnut combs. These are then crossed to each other and the F2 progeny are: 93 walnut, 11 single, 29 rose, 32 pea.

A) How is comb morphology inherited?

B) What would the progeny be if you crossed a homozygous rose combed hen with a homozygous pea combed roster? What are the phenotypic ratios that you would see in the F2 progeny?

C) If a walnut rooster is crossed to a pea hen and the progeny are 12 walnut, 11 pea, 3 rose and 4 single, what are the genotypes of the parents?

D) If a walnut rooster is crossed to a rose hen and all the progeny are walnut, what are the genotypes of the parents?

Answer 1

a) there are two genes that control the comb morphology.

let the genes be R and P.

R_pp- rose comb, rrP_ - pea comb, R_P_ - walnut comb, rrpp- single comb.

B) the genotype of the  homozygous rose combed hen is RRpp and the genotype of the homozygous pea combed rooster is rrPP

RRpp * rrPP

RrPp ( walnut comb)

RrPp * RrPp

	RP	Rp	rP	rp
RP	RRPP ( walnut)	RRPp( walnut)	RrPP ( walnut)	RrPp ( walnut)
Rp	RRPp( walnut)	RRpp( rose)	RrPp ( walnut)	Rrpp ( rose)
rP	RrPP( walnut)	RrPp ( walnut)	rrPP ( pea)	rrPp ( pea)
rp	RrPp( walnut)	Rrpp ( rose)	rrPp (pea)	rrpp ( single)

the phenotypic ratio is

walnut: rose: pea: single=9:3:3:1

C) when Walnut rooster is crossed to pea combed hen all four phenotypes are possible so the walnut rooster has all four alleles and the pea combed hen is heterozygous at the P locus

RrPp * rrPp

	RP	Rp	rP	rp
rP	RrPP( walnut)	RrPp ( walnut)	rrPP ( pea)	rrPp ( pea)
rp	RrPp( walnut)	Rrpp ( rose)	rrPp (pea)	rrpp ( single)

the genotypes of the parents are

Pea combed hen: rrPp

walnut combed rooster: RrPp.

D) all the progenies are walnut so the genotype of the walnut rooster is RRPP and the genotype of the rose hen is R_pp

RRPP * Rrpp

	Rp	rp
RP	RRPp (walnut)	RrPp(walnut)
RP	RRPp(walnut)	RrPp(walnut)